What torque would have to be applied to a rod with a length of $6 m$ $6 m$ and a mass of $3 k g$ $3 kg$ to change its horizontal spin by a frequency $3 H z$ $3 Hz$ over $8 s$ $8 s$ ?

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Answer 1 · 2017-01-16T15:14:35

The torque is $21.2 N m$ $21.2Nm$

The torque is the rate of change of angular momentum

$τ = \frac{d L}{d t} = \frac{d (I ω)}{d t} = I \frac{d ω}{d t}$ $tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt$

The moment of inertia of a rod is $I = \frac{1}{12} \cdot m L^{2}$ $I=1/12*mL^2$

$= \frac{1}{12} \cdot 3 \cdot 6^{2} = 9 k g m^{2}$ $=1/12*3*6^2= 9 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{3}{8} \cdot 2 π$ $(domega)/dt=(3)/8*2pi$

$= (\frac{3 π}{4}) r a d s^{- 2}$ $=((3pi)/4) rads^(-2)$

So the torque is $τ = 9 \cdot \frac{3 π}{4} N m = \frac{27}{4} π N m = 21.2 N m$ $tau=9*(3pi)/4 Nm=27/4piNm=21.2Nm$

What torque would have to be applied to a rod with a length of 6m6 m and a mass of 3kg3 kg to change its horizontal spin by a frequency 3Hz3 Hz over 8s8 s?